orbits of a normal subgroup are equal in size when the full group acts transitively

The following theorem proves that if a group acts transitively on a finite set, then any of the orbits of any normal subgroup are equal in size and the group acts transitively on them. We also derive an explicit formula for the size of each orbit and the number of orbits.

Theorem 1.

Let H be a normal subgroup of G, and assume G acts transitively on the finite set A. Let O1,…,Or be the orbits of H on A. Then

1.

G permutes the 𝒪i transitively (i.e. for each g∈G,1≤j≤r, there is 1≤k≤r such that g⁢𝒪j=𝒪k, and for each 1≤j,k≤r, there is g∈G such that g⁢𝒪j=𝒪k), and the 𝒪i all have the same cardinality.

2.

If a∈𝒪i, then |𝒪i|=|H:H∩Ga| and r=|G:HGa|.

Proof.

Note first that if g∈G,a∈𝒪i, and g⋅a∈𝒪j, then g⁢𝒪i⊂𝒪j. For suppose also b∈𝒪i. Then since a,b are in the same H-orbit, we can choose h∈H such that b=h⋅a. Then

g⋅b=g⋅h⋅a=g⋅h⋅g-1⋅g⋅a=h′⋅g⋅a∈h′⁢𝒪j⊂𝒪j

since H is normal in G. Thus for each g∈G,1≤j≤r, there is 1≤k≤r such that g⁢𝒪j⊂𝒪k.

Given j,k, choose aj∈𝒪j,ak∈𝒪k. Since G is transitive on A, we may choose g∈G such that g⋅aj=ak. It follows from the above that g⁢𝒪j⊂𝒪k.

To prove 1), given j,k, choose g such that g⁢𝒪j⊂𝒪k and g′ such that g′⁢𝒪k⊂𝒪j. But then |𝒪j|≤|𝒪k|≤|𝒪j| so that |𝒪j|=|𝒪k| and the subset relationships in the previous two paragraphs are actually set equality.